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jarptica [38.1K]
1 year ago
12

Solve the following inequality for s. Write your answer in simplest form.

Mathematics
2 answers:
Doss [256]1 year ago
3 0

\qquad\qquad\huge\underline{{\sf Answer}}♨

Let's solve ~

\qquad \sf  \dashrightarrow \: - 10 + 5(78 + 8) <  - 4s + 3 - 8

\qquad \sf  \dashrightarrow \: - 10 + 5(86) <  - 4s  - 5

\qquad \sf  \dashrightarrow \: - 10 +430 <  - 4s  - 5

\qquad \sf  \dashrightarrow \: 420 + 5 <  - 4s

\qquad \sf  \dashrightarrow \: 42 5 <  - 4s

\qquad \sf  \dashrightarrow \: 42 5 \div 4 <   - s

\qquad \sf  \dashrightarrow \: 106.25 <   - s

Inequalities change when it is multiplied by -ve number

\qquad \sf  \dashrightarrow \:  - 106.25  >    s

Mumz [18]1 year ago
3 0

\qquad \sf  \: - 10 + 5(78 + 8) < - 4s + 3 - 8

\qquad \sf   \: - 10 + 5(86) < - 4s - 5

\qquad \sf   \: - 10 +430 < - 4s - 5

\qquad \sf   \: 420 + 5 < - 4s

\qquad \sf   \: 42 5 < - 4s

\qquad \sf  \: 42 5 \div 4 < - s

\qquad \sf \: 106.25 < - s

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1 year ago
Kermit's favorite iced tea is made with 151515 tea bags in every 222 liters of water. Peggy made a 121212-liter batch of iced te
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Answer:

The answer is (Choice C)  It is just right.

Step-by-step explanation:

<u><em>The question is incomplete, so the complete question is below:</em></u>

Kermit's favorite iced tea is made with 15 tea bags in every 2 liters of water. Peggy made a 12-liter batch of iced tea with 90 tea bags. What will Kermit think of Peggy's iced tea? Choose 1 answer: (Choice A)  It is too strong. (Choice B)  It is too weak. (Choice C)  It is just right.

Now, to find which of the choice is correct.

So, we find first the unit rate of Peggy's iced tea bags per liter:

If, Peggy made a 12-liter batch of iced tea with 90 tea bags.

Then, Peggy would made a 1-liter batch of iced tea with = \frac{90}{12} =7.5\ tea\ bags.

Now, to get the unit rate of Kermit's iced tea bags per liter:

If, Kermit made a 2 liter of iced tea with 15 tea bags.

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Answer:

The median of A is the same as the median of B.

The interquartile range of B is greater than the interquartile range of A.

Step-by-step explanation:

Given that:

A = number of runs allowed in first 9 games

A = {1, 4, 2, 2, 3, 1, 1, 2, 1}

Rearranging A : 1, 1, 1, 1, 2, 2, 2, 3, 4

Median A = 1/2(n + 1) th term

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Number of runs allowed in 10th game = 9

B = {1, 4, 2, 2, 3, 1, 1, 2, 1, 9}

Rearranging B = 1, 1, 1, 1, 2, 2, 2, 3, 4, 9

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Interquartile range = Q3 - Q1 = 3.5 - 1 = 2.5

Median A = 2 ; median B = 2

IQR B = 2.5 ; IQR A = 1.5 ; IQR B > IQR A

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